Verified answer

How about letting t = x^(-1/2)

dt = -1/2 * x^(-3/2) * dx

So integral of lnx * x^(-3/2) * dx

= -2 * ln(1/t^2) * dt

= -2 * ln[t^-2] * dt

= 4*lnt*dt

Now you can integrate by parts using

u = ln(t), dv = 4

du = dt/t, v = 4t

integral = 4t*ln(t) - integral of 4*dt

= 4*t*[ln(t) - 1]

= 4/sqrt(x) * [ln(x^-1/2) - 1]

= 4 * [-1/2 * lnx - 1] / sqrt(x)

= -2 * [lnx - 2 ] / sqrt(x)

Hello

[(1/x)*(x^(3/2)) - (3/2)x^(1/2)*lnx]/x^3

You will need to simply this.

Hope This Helps!

why, it shoud:

u = ln x

x = exp(u)

int of (lnx) / x^3/2 dx =

int of u / exp(u)^3/2 d[exp(u)] =

int of u / exp(u)^3/2 exp(u) du =

int of u / exp(u)^1/2 du =

int of u exp(-u/2) du =

-2 int of u d[exp(-u/2)] =

by part:

-2( u exp(-u/2) - int of exp(-u/2) du) =

-2(u exp(-u/2) + 2 exp(-u/2)) + C=

(4 -2u)exp(-u/2) + C=

(4 -2ln x)exp(-ln x/2) + C

(4 -2ln x)/√exp(ln x) + C

(4 -2ln x)/√x + C